Answer: The mass of carbon dioxide produced in the given reaction is 132 grams.
Step-by-step explanation:
To calculate the number of moles, we use the equation:
.....(1)
Given mass of ethyne = 38.9 g
Molar mass of ethyne = 26 g/mol
Putting values in equation 1, we get:
![\text{Moles of ethyne}=(38.9g)/(26g/mol)=1.5mol](https://img.qammunity.org/2020/formulas/chemistry/middle-school/phva5w5mj5lm0ntb11l7per9rbik8gnt6u.png)
- The chemical reaction for the combustion of ethyne follows the equation:
![2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O](https://img.qammunity.org/2020/formulas/chemistry/middle-school/ovxt33vyhnnnzjv1un24oz50pricrab2ky.png)
By Stoichiometry of the reaction:
2 moles of ethyne produces 4 moles of carbon dioxide.
So, 1.5 moles of ethyne will produce =
of carbon dioxide.
- Now, calculating the mass of carbon dioxide from equation 1, we get:
Molar mass of carbon dioxide = 44 g/mol
Moles of carbon dioxide = 3 moles
Putting values in equation 1, we get:
![3mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=132g](https://img.qammunity.org/2020/formulas/chemistry/middle-school/xlu8h65ewo5oc9m7kzhtoq117bnfut60qf.png)
Hence, the mass of carbon dioxide produced in the given reaction is 132 grams.